# Co-algebra

2010 Mathematics Subject Classification: *Primary:* 16T15 [MSN][ZBL]

A module $A$ over a commutative ring $k$ with two homomorphisms, "comultiplication" $\phi : A \to A \otimes_k A$ and "counit" $\epsilon : A \to k$ such that the diagrams $$ \begin{array}{ccc} A & \stackrel{\phi}{\rightarrow} & A \otimes A \\ {}^\phi\downarrow & \ & \downarrow{}^{1 \otimes \phi} \\ A \otimes A & \stackrel{\phi \otimes 1}{\longrightarrow} & A \otimes A \end{array} $$ and $$ \begin{array}{ccccc} A \otimes A & \stackrel{\phi}{\leftarrow} & A & \stackrel{\phi}{\rightarrow} & A \otimes A \\ & & & & \\ & \searrow{}^{\epsilon\otimes1}\ & \Vert & {}^{1\otimes\epsilon}\swarrow & \\ & & & & \\ & & A & & \end{array} $$ are commutative. In other words, a co-algebra is the dual concept (in the sense of category theory) to the concept of an associative algebra over a ring $k$.

Co-algebras have acquired significance in connection with a number of topological applications such as, for example, the simplicial complex of a topological space, which is a co-algebra. Closely related to co-algebras are the Hopf algebras, which possess algebra and co-algebra structures simultaneously (cf. Hopf algebra).

#### References

[1] | S. MacLane, "Homology" , Springer (1963) |

#### Comments

Given a co-algebra $A$ over $k$, let $A^*$ be the module of $k$-module homomorphisms from $A$ to $k$. For $f,g \in A^*$ define the product $fg : A \to k$ by the formula $fg : a \mapsto (f\otimes g)(\phi(a))$, where $k \otimes_k k$ is identified with $k$. For any two $k$-modules $M,N$ define $\rho : M^* \otimes N^* \to (M \otimes N)^*$ by $\rho(f\otimes g)(m\otimes n) = f(m)g(n)$. Then the multiplication on $A^*$ can also be seen as the composite $A^* \otimes A^* \to (A\otimes A)^* \stackrel{\phi^*}{\to} A^*$. The element $\epsilon : A \to k$ is a unit element for this multiplication making $A^*$ an associative algebra with unit, the dual algebra. In general the mapping $\rho$ is not an isomorphism and there is no natural $k$-module homomorphism $M^* \otimes N^* \to (M \otimes N)^*$. Thus there is no equally natural construction associating a co-algebra to an algebra over $k$, even when $k$ is a field. In that case there does however exist an adjoint functor $A \mapsto A^0$ to the functor $C \to C^*$ which associates to a co-algebra its dual algebra, i.e. $\textsf{Alg}_k (A,C^*) \sim \textsf{Coalg}_k(A^0,C)$ for $A \in \textsf{Alg}_k$, $C \in \textsf{Coalg}_k$, where $\textsf{Alg}_k$ and $\textsf{Coalg}_k$ denote, respectively, the category of $k$-algebras and the category of $k$-co-algebras, [a2]; cf. also Hopf algebra. But if $B$ is free of finite rank over $k$ then $\rho : B^* \otimes B^* \to (B \otimes B)^*$ is an isomorphism and the dual co-algebra can be defined.

Let $S$ be the set $s_0, s_1, \ldots$. Let $kS = \oplus_i ks_i$ and define $$ \phi(s_n) = \sum_{i-0}^n s_i \otimes s_{n-i}\,,\ \ \ \epsilon(s_i) = 1 \ . $$

Then $kS$ is a co-algebra.

If $(A,\phi)$ and $(B,\psi)$ are two co-algebras, then a morphism of co-algebras is a $k$-module morphism $\alpha:A \to B$ such that $\psi \circ \alpha = (\alpha\otimes\alpha)\circ\phi$ and $\epsilon_B \circ \alpha = \alpha \circ \epsilon_A$. A co-ideal of a co-algebra $A$ is a $k$-submodule $V$ such that $\phi(V) \subset V \otimes A + A \otimes V$ and $\epsilon(V) = 0$.

A co-module $M$ over a co-algebra $(C,\phi)$ is a $k$-module with a $k$-module morphism $\psi:M \to M\otimes C$ such that $(\psi\otimes1)\circ\psi = (1\otimes\phi)\circ\psi$ and $(1\otimes\epsilon)\circ\psi$ the canonical isomorphism $M \to M \otimes k$. There are obvious notions of homomorphisms of co-modules, etc.

#### References

[a1] | M. Sweedler, "Hopf algebras" , Benjamin (1969) Zbl 0194.32901 |

[a2] | E. Abe, "Hopf algebras" , Cambridge Univ. Press (1980) ISBN 0-521-60489-3 Zbl 0476.16008 |

**How to Cite This Entry:**

Co-algebra.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Co-algebra&oldid=40943